I claim that there is a 'path' connecting $k$ and $1$.

Question

Let $S=\lbrace 0,1,2,...,n\rbrace$. Define $S_0=S-\lbrace 0\rbrace$ and $S_k=S-\lbrace k\rbrace$ for some $k\in S_1$. Let $f:S_0\longrightarrow S_k$ be such that $f(i)=(k+i)$ mod $n+1$. I claim that there is a 'path' connecting $k$ and $0$. i.e $f(k)=,...,=0$.

I have a sketchy 'proof' but I'm not sure I've done it carefully.

Example: $S=\lbrace 0,1,2,3,4\rbrace$. $S_0=\lbrace 1,2,3,4\rbrace$ and $S_3=\lbrace 0,1,2,4\rbrace$. So $f(1)=4$, $f(2)=0$, $f(3)=1$, $f(4)=2$. The path is $f(3)=1$, $f(1)=4$, $f(4)=2$, $f(2)=0$.

Answer 1

If you think of your set as the group $G=\mathbb{Z}/(n+1)$, then $f$ is (when extended to include zero) the map $f:G\to G:x\mapsto x+k$. Then $f^{i}(k) = (i+1)k$, so that $f^n(k) = 0$.